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NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:49 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 5 covers Complex numbers and its applications. Complex numbers are something which we have been studying over the years with any ‘real number’ being said to be a complex number, which can be denoted as a variable or alphabet in simpler terms. The other number, along with the real number, is a part of the complex number, is called the ‘imaginary number.’ Both of them appear together in a problem. Terms such as integer, conjugate, square root, polar, etc. will be studied in NCERT Exemplar Class 11 Maths solutions chapter 5, that will be utilised to find results for these complex number problems.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Suggested: JEE Main: high scoring chaptersPast 10 year's papers

This Story also Contains
  1. NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3
  2. Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5
  3. Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5
  4. What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?
  5. NCERT Solutions for Class 11 Mathematics Chapters
  6. Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3

Question:1

For a positive integer n, find the value of (1i)n (11/i)n

Answer:

(1i)n(11/i)n
=(1i)n(1+i)n
=(1i2)n
=2n

Question:2

Evaluate i=113(in+in+1) where n ϵN.

Answer:

= i=113(in+in+1)
= i=113(1+i)in
=(1+i)(1+i2+i3+i4....+i12+i13)
=(1+i)i(i131)(i1)
=(1+i)i(i1)(i1)
=(1+i)i
=i+i2
=i1

Question:3

If (1+i)(1i)3(1i)(1+i)3=x+iythen find (x, y).

Answer:

Given (1+i1i)3(1i1+i)3=x+iy
=(1+i1i)3(1i1+i)3
=(1+2i+i21i2)3(12i+i21i2)3
=(2i2)3(2i2)3
=i3(i3)
=2i3=02i
Thus,(x,y)=(0,2)

Question:4

If (1+i)2/(2i)=x+iy then find the value of x + y.

Answer:

x+iy=(1+i)2(2i)=(1+2i+i2)(2i)=2i(2i)

Onrationalising2i(2+i)/(2i)(2+i),
=(4i+2i2)(4i2)=((4i2)(4+1))
=25+4i5
x=25andy=45

So,x+y=25+45=25

Question:5

If (1i1+i)100=a+ib then find (a, b).

Answer:

a+ib=(1i1+i)100=[1i1+i1i1i]100
=[(1i)21i2]100
=((12i+i2)(1+1))100
=(2i2)100
=(i4)25=1
Hence,(a,b)=(1,0)

Question:6

If a=cosθ+isinθ, find the value of(1+a)(1a) .

Answer:

a=cosθ+isinθ
(1+a)(1a)=(1+cosθ)+isinθ(1cosθ)isinθ

=(2cos2θ2+i2sinθ2cosθ2)(2sin2θ2i2sinθ2cosθ2)
=2cosθ2(cosθ2+isinθ2)2sinθ2(sinθ2icosθ2)
=icosθ2(cosθ2+isinθ2)sinθ2(isinθ2i2cosθ2)
=icosθ2(cosθ2+isinθ2)sinθ2(isinθ2+cosθ2)
=icotθ2

Question:7

If (1+i)z=(1i)z¯, then show that z=iz¯

Answer:

(1+i)z=(1i)z¯,
z=1i1+iz¯
Rationalising the denominator,
(1i)(1i)(1+i)(1i)z¯
=(1i)21i2z¯
=(12i+i2)1+1z¯
=(12i1)2z¯
=iz¯
Hence,Proved.

Question:8

Ifz=x+iy, then show that zz¯+2(z+z¯)+b=0 where b?R, representing z in the complex plane is a circle.

Answer:

z=x+iy
z¯=xiy
Now,wealsohave,zz¯+2(z+z¯)+b=0
(x+iy)(xiy)+2(x+iy+xiy)+b=0
This equation is a equation of circle

Question:9

If the real part of (z¯+2)(z¯1) is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

Letz=x+iy
z¯+2z¯1=xiy+2xiy1
=[(x+2)iy][(x1)+iy][(x1)iy][(x1)+iy]
=(x1)(x+2)+y2+i[(x+2)y(x1)y](x1)2+y2
realpart=4(x1)(x+2)+y2(x1)2+y2=4
x2+x2+y2=4(x22x+1+y2)
3x2+3y29x+6=0

The equation obtained is that of a circle. Hence, locus of z is a circle.

Question:10

Show that the complex number z, satisfying the condition arg(z1z+1)=π4 lies on a circle.

Answer:

Let z=x+iy
arg(z1z+1)=π4
arg(z1)arg(z+1)=π4
arg(x+iy1)arg(x+iy+1)=π4
arg(x1+iy)arg(x+1+iy)=π4
tan1yx1tan1yx+1=π4
tan1yx1yx+11+(yx1)(yx+1)=π4
tan1(y(x+1+x+1)x21+y2)=π4
tanπ4=2yx2+y21
x2+y21=2y
x2+y212y=0
The equation obtained represents the equation of a circle

Question:11

Solve that equation |z|=z+1+2i.

Answer:

|z|=z+1+2i
Substitutingz=x+iyweget|x+iy|=x+iy+1+2i
|z|=(x2+y2)=(x+1)+i(y+2)
Comparing real and imaginary parts (x2+y2)=(x+1)
And0=y+2,y=2
Substituting the value of y in(x2+y2)=(x+1)
x2+(2)2=(x+1)2
x2+4=x2+2x+1

Hence, x=32
Hence,z=x+iy
=322i

Question:12

If |z+1|=z+2(1+i), then find z

Answer:

We have |z+1|=z+2(1+i)
Substituting z=x+iy we get x+iy+1=x+iy+2i+1
|z|=(x2+y2)=(x+1)2+y2=(x+2)+i(y+2)
Comparing real and imaginary parts, (x+1)2+y2=(x+2)
and0=y+2;y=2
Substituting the value of y in (x+1)2+y2=(x+2)
(x+1)2+(2)2=(x+2)2
x2+2x+1+4=x2+4x+4
2x=1
Hence, x=12
Thus, z=x+iy=122i

Question:13

If arg(z1)=arg(z+3i), then find x – 1 : y. where z=x+iy

Answer:

Given that arg(z1)=arg(z+3i)
arg(x+iy1)=arg(x+iy+3i)
arg(x1+iy)=arg(x+i(y+3))
tan1yx1=tan1y+3x
yx1=y+3x
xy=xyy+3x3
3x3=y
x1y=13

Question:14

Show that |z2z3|=2 represents a circle. Find its centre and radius.

Answer:

|z2z3|=2 Substituting z=x+iy, we get |x+iy2x+iy3|=2
|x2+iy|=2|x3+iy|
(x2)2+y2=2((x3)2+y2
x24x+4+y2=4(x26x+9+y2)
3x2+3y220x+32=0
x2+y2203x+323=0
(x103)2+y2+3231009=0
Thus, the centre of circle is (103,0) and radius is 2/3

Question:15

If z1z+1 is a purely imaginary number(z1), then find the value of |z|.

Answer:

Let z=x+iy
z1z+1=x+iy1x+iy+1
[(x1)+iy][(x+1)iy][(x+1)+iy][(x+1)iy]
=(x1)(x+1)+y2+i[(x+1)y(x1)y](x+1)2+y2
z1z1 is purely imaginary
(x1)(x+1)+y2(x+1)2+y2=0
x21+y2=0
x2+y2=1 Hence, |z|=1

Question:16

z1 and z2 are two complex numbers such that |z1|=|z2| and arg(z1)+arg(z2)=π, then show that |z1|=|z2| .

Answer:

Let z1=|z1|(cosθ1+i sinθ1) and z2=|z2|(cosθ2+i sinθ2)
Given that |z1|=|z2|
And arg(z1)+arg(z2)=π

θ1+θ2=π

θ1=πθ2

z2=|z2|(cosθ2+i sinθ2)

z1=|z2|(cosθ2+isinθ2)

z1=|z2|(cosθ2isinθ2)

z1=|z2|(cosθ2isinθ2), |z1|=|z2|

Question:17

If |z1| = 1 (z1 ≠ –1) and z2=z11z1+1 then show that the real part of z2 is zero

Answer:

Let z1=x+iy |z1|=x2+y2=1
z2=z11z1+1=x+iy1x+iy+1
=[(x1)+iy][(x+1)iy][(x+1)+iy][(x+1)iy]
=(x1)(x+1)+y2+i[(x+1)y(x1)y](x+1)2+y2
x2+y21+2iy(x+1)2+y2=0
Since, x2+y2=1
11+2iy(x+1)2+y2=0+2iy(x+1)2+y2
Therefore, the real part of z2 is zero

Question:18

If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find arg(z1z4)+arg(z2z3)

Answer:

z1 and z2 are conjugate complex numbers.
The negative side of the real axis =r1(cosθisinθ)
=r1(cos(θ1)+isin(θ1))
Similarly, z3=r2(cos(θ2)isin(θ2))
z4=r2(cos(θ2)+isin(θ2))
arg(z1z4)+arg(z2z3)=arg(z1)arg(z4)+arg(z2)arg(z3)??
=θ1(θ2)+(θ1)θ2=0

Question:19

If |z1|=|z2|=...=|zn|=1, then
Show that |z1+z2+z3+...+zn|=|1/z1+1/z2+...+1/zn|

Answer:

|z1|=|z2|=...=|zn|=1
|z1|2=|z2|2=...=|zn|2=1
z1z1¯=z2z2¯=...=znzn¯
z1=1z1¯,z2=1z2¯,...zn=1zn¯
Now, |z1+z2+z3+...zn| =|z1z1¯z1¯+z2z2¯z2¯+...+znzn¯zn¯|
=|1z1¯+1z2¯+...+1zn¯|
=|1z1+1z2+...+1zn|

Question:20

If for complex numbers z1 and z2, arg(z1)arg(z2)=0, then show that |z1z2|=|z1||z2|

Answer:

Let z1=|z1|(cosθ1+isinθ1) and z2=|z2|(cosθ2+isinθ2)
arg(z1)arg(z1)=0
θ1θ2=0
θ1=θ2
z1z2=(z1|cosθ1|z2|cosθ1)+i(z1|sinθ1|z2|sinθ1)
z1z2=(z1|cosθ1|z2|cosθ1)2+(z1|sinθ1|z2|sinθ1)2
=|z1|2+|z2|22|z1||z2|cos2θ12|z1||z2|sin2θ1
=|z1|2+|z2|22|z1||z2|
=(|z1||z2|)2=|z1||z2|

Question:21

Solve the system of equations Re(z2)=0,|z|=2.

Answer:

Re(z2)=0,|z|=2.
Letz=x+iy
|z|=x2+y2=2
x2+y2=4
z2=x2+2ixyy2
=(x2y2)+2ixy
Now,Re(z2)=0
x2y2=0
x2=y2=2
x=y=±2
Hence, z=x+iy=±2±i2
=2+i2,2i2,2+i2and2i2

Question:22

Find the complex number satisfying the equation z+2|(z+1)|+i=0 .

Answer:

z+2|(z+1)|+i=0...(i)
Substituting z=x+iy we get x+iy+2|x+iy+1|+i=0
x+i(1+y)+2[(x+1)2+y2]=0
x+i(1+y)+2[(x2+2x+1+y2)]=0
1+y = 0
y = -1
x+2x2+2x+2=0
2x2+2x+2=x
2x2+4x+4=x2
x2+4x+4=0
(x+2)2=0
x=2
Hence,z=x+iy=2i

Question:23

Write the complex number z=1icosπ3+isinπ3 in polar form

Answer:

z=1icosπ3+isinπ3
=2[12i1/2]cosπ3+isinπ3
=2[cosπ4isinπ4]cosπ3+isinπ3
=2[cos(π4π3)+isin(π4π3)]
=2[cos(7π12)+isin(7π12)]
=2[cos(5π12)+isin(5π12)]

Question:24

If z and w are two complex numbers such that |zw|=1 and arg(z)arg(w)=π/2, then show that zw=i.

Answer:

Let z=|z|(cosθ1+isinθ1) and w=|w|(cosθ2+isinθ2)
|zw|=|z||w|=1
Also arg(z)arg(w)=π2
θ1θ2=π2
z¯w=|z|(cosθ1isinθ1)w
w=|w|(cosθ2+isinθ2)=1
z¯w=|z||w|(cos(θ1)+isin(θ1))(cosθ2+isinθ2)
z¯w=cos[(θ2θ1)+isin(θ2θ1)]
=[cos(π2)+isin(π2)]
=1[0i]=i

Question:25

i) For any two complex numbers z1, z2 and any real numbers a, b, |az1bz2|2+|bz1+az2|2=....

ii) The value of 259 is

iii) The number (1i)31i3 is equal to .......

iv) The sum of the series i+i2+i3+...+i1000 upto 1000 terms is ...

v) Multiplicative inverse of 1 + i is ................

vi) If z1 and z2 are complex numbers such that z1+z2 is a real number, then z2 =

vii) arg(z)+arg(z¯) (z¯0) is ....

viii) If |z+4|3then the greatest and least values of |z+1| are ..... and .....

ix) if z2z+2=π6then the locus of z is .......

x) If |z|=4 and arg(z)=5π6 then z=

Answer:

i) |az1bz2|2+|bz1+az2|2
=|az1|2+|bz2|22Re(az1.b(z2¯)+|bz1|2+|az2|2+2Re(az1.bz2¯)?)
=|az1|2+|bz2|2+|bz1|2+|az2|2=(a2+b2)(|z1|2+|z2|2)
ii)259=5i3i=15i2=15
iii) (1i)31i3=(1i)3(1i)(1+i+i2)
=(1i)3(1+i1)=1+i22ii
=112ii=2ii=2
iv) i+i2+i3+...+i1000=0[n=11000in=0]
v) 11+i=11i(1+i)(1i)=12(1i)
vi) Let z1=x1+iy1andz2=x2+iy2
z1+z2=(x1+x2)+i(y1+y2)
If z1+z2 is real then y1+y2=0
y1=y2
z2=x2iy1
z2=x1iy1(whenx1=x2)
So, z2=z1¯
vii) arg(z)+arg(z¯)
Ifarg(z)=θ,thenarg(z¯)=θ
θ+(θ)=0
viii) |z+4|3
=|z+43||z+4|+|3|
=|z+43|3+3
=|z+43|6
The gratest value is 6 and the least value of |z+1| is 0
ix)z2z+2=π6
Let z=x+iy
|x+iy2x+iy+2|=|(x2)+iy(x+2)+iy|=π6
6|(x2)+iy|=π|(x+2)+iy|
6(x2)2+y2=π(x+2)2+y2
36[x2+44x+y2]=π2[x2+4+4x+y2]
36x2+144144x+36y2=π2x2+4π2+4π2x+π2y2
(36π2)x2+(36π2)(144+4π2)x+1444π2=0
which represents an equation of a circle
x) |z|=4 and arg(z)=5π6
Let z=x+iy
|z|=x2+y2=4
x2+y2=16
arg(z)=tan1yx=5π6
yx=tan(ππ6)
=tanπ6=13
x=3y
(3y)2+y2=16
3y2+y2=16
4y2=16
y2=4
y=±2
x=23
z=23+2i

Question:26

State True of False for the following:

i) The order relation is defined on the set of complex numbers.

ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z2) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z1 and Z2 be two complex numbers such that |z, + z2| = |z1 j + |z2|, then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex number can be compared. So, it is false.

(ii) Let z=x+iy

z.i=(x+iy)i=xi=y which rotates at angle of 180. So, it is ‘false’.

(iii) Let z=x+iy

|z|+|z1|=x2+y2+(x1)2+y2

The value of |z|+|z1| is minimum when x=0,y=0i.e.,1

Hence, it is true.

iv) Let z=x+iy

Given that |z1|=|zi|

|x+iy1|=|x+iyi|

|(x1)+iy|=|x(1y)i|

(x1)2+y2=x2+(1y)2

(x1)2+y2=x2+(1y)2

x22x+1+y2=x2+1+y22y

2x+2y=0xy=0 which is a straight line slope=1

Now, equation of line through the point 1,0and 0,1

y0=1001(x1)

y=x+1 whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let z=x+iy z0andRe(z)=0

Since, real part is 0 x=0,z=0+iy=iy

1m(z2)=y2i2=y2which is real Hence, it is False.

vi) (z4)<|z2|

Let z=x+iy

|x+iy4|<|x+iy2|

|(x4)+iy|<|(x2)+iy|

(x4)2+y2<(x2)2+y2

(x4)2+y2<(x2)2+y2

(x4)2<(x2)

x2+168x<x2+44x

8x+4x<16+4

4x<12

x>3 Hence, it is true.

vii) Let z1=x1+iy1 and z2=x2+iy2

|z1+z2|=|z1|+|z2|

|x1+iy1+x2+iy2|=|(x1+iy1)+(x2+y2i)|

=(x1+x2)2+(y1+y2)2

=x12+y12+x22+y22

Squaring both sides, we get (x12+x22+2x1x2+y12+y22+2y1y2)

=x12+y12+x22+y22+2x12x22+x12y22+x22y12+y12y22

=2x1x2+2y1y2=x12x22+x12y22+x22y12+y12y22

=x1x2+y1y2=x12x22+x12y22+x22y12+y12y22

Again squaring on both sides we get x12x22+x12y22+2x1y1x2y2=x12x22+x12y22+x22y12+y12y222x1y1x2y2=x12y22+x22y12

x12y12+x22y122x1y1x2y2=0

(x1y2x2y1)2=0

x1y2x2y1=0

x1y2=x2y1

x1y1=x2y2

y1x1=y2x2

arg(z1)=arg(z2)=0 Hence, it is false.

(viii) Since, every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question:27

COLUMN A

COLUMN B

a) The polar form of i+3 is

i) Perpendicular bisector of segment joining (-2,0) and (2,0)

b)The amplitude of 1+3 is

ii) On or outside the circle having centre at (0,-4) and radius 3.

c) If |z+2|=|z2| then locus of z is

iii) 2π/3

d)If |z+2i|=|z2i| then locus of z is

iv)Perpendicular bisector of segment joining (0,-2) and (0,2)

e) Region represented by |z+4i|3 is

v)2(cosπ6+isinπ6)

f) Region represented by |z+4|3 is

vi) On or outside the circle having centre at (-4,0) and radius 3units.

g)Conjugate of 1+2i1i lies in

vii) First Quadrant

h) Reciprocal of 1-i lies in

viii) Third Quadrant

Answer:

a)Given z=i+3

Polar form of z =r[cosθ+isinθ]=i+3.

r=3+1=2

And tanα=13

α=π6

Since x>0,y>0

Polar form of z =2[cosπ6+isinπ6]

b) Given that z=1+3=1+3i

Here argument z =tan1|31|=tan13=π3

So, α=π3

Since, x<0 and y>0 θ=πα=ππ3=2π3

c) |z+2|=|z2|

|x+iy+2|=|x+iy2|

|(x2)+iy|=|(x2)+iy|

(x+2)2+y2=(x2)2+y2

(x+2)2+y2=(x2)2+y2

(x+2)2=(x2)2

x2+4+4x=x2+44x

8x=0,x=0

which represents equation of-y axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) |z+2i|=|z2i|

(x+iy+2i)=|x+iy2i|

|x+(y+2)i|=|x+(y2)i|

x2+(y+2)2=x2+(y2)2

x2+(y+2)2=x2+(y2)2

(y+2)2=(y2)2

y2+4+4y=y2+44y

8y=0,y=0

which is the equation of x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) |z+4i|3

|x+iy+4|3

|x+(y+4)i|3

x2+(y+4)23

x2+(y+4)29

x2+y2+8y+169

x2+y2+8y+70 r=(4)27=3

which represents a circle on or outside having centre (0,-4)

f) |z+4|3

Let z=x+iy Then, |x+iy+4|3

|(x+4)+iy|3

(x+4)2+y23

x2+8x+16+y29

x2+y2+8x+70 which is a circle having centre (-4,0) and r=(4)27=9=3 and is on the circle.

g) Let z=1+2i1i

1+2i1i1+i1+i=1+i+2i+2i21i2

=1+i+2i21+1=1+3i2

=1+i+2i21+1=1+3i2

=12+32i which lies in third quadrant..

h) Given that z=1-i

Reciprocal of z=1z=11i1+i1+i

=1+i1+i2 which lies in first quadrant.

So,(a)(v),(b)(iii),(c)(i),(d)(iv),(e)(ii),(f)(vi),(g)(viii),(h)(vii)

Question:28

What is the conjugate of 2i(12i)2?

Answer:

2i(12i)2=2i1+4i24i
=2i34i=2i34i3+4i3+4i
=6+8i+3i4i2916i2
6+11i+4916i2=2+11i9+16
2+11i25=225+1125i
Hence, z¯=2251125i

Question:29

If |z1|=|z2|, is it necessary that z1=z2?

Answer:

Let z1=x1+iy1andz2=x2+iy2
|x1+iy1|=|x2+iy2|
=x12+y12=x22+y22
x12+y12=x22+y22
Consider z1=3+4iz2=4+3i
z1z2 Hence, it is not necessary that z1=z2

Question:31

Find z if |z|=4 and arg(z)=5π/6..

Answer:

Polar form of z=r[cosθ+isinθ]
=4[cos5π6+isin5π6]
=4[cos(ππ6)+isin(ππ6)]
=4[cos(π6)+isin(π6)]
=4[32+i12]
=23+2i

Question:32

Find (1+i)(2+i)(3+i).

Answer:

|(1+i)(2+i)(3+i)3i3i|=|(1+i).62i+3ii29i2|
=|(1+i)(7+i)9+1|
=|7+i+7i+i29+1|
=|7+i+7i+i210|
=|7+8i110|
=|6+8i10|
=|35+45i|
=(35)2+(45)2=1

Question:33

Find principal argument of (1+i3)2.

Answer:

(1+i3)2=1+i23+23i=13+23i=2+23itanα=|232|tanα=|3|=3tanα=tanπ3α=π3

Now Re(z)<0 and image(z)>0 arg(z)=πα=ππ3=2π3
Hence, the principal arg=2π3

Question:34

Where does z lie, if |z5iz+5i|=1.

Answer:

 Let z=x+iy|x+yi5ix+iy+5i|=1|x+(y5)i|=|x+(y+5)i|x2+(y5)2=x2+(y+5)2(y5)2=(y+5)2y2+2510y=y2+25+10y20y=0y=0

Hence, z lies on x-axis i.e., real axis.


Question:35

sinx+icos2x and cosxisin2x are conjugate to each other for:
A. x=nπ

B. x=(n+12)π2

C. x=0
D. No value of x

Answer:

The answer is the option (c).
\text{let }z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x

But \: \: we \: \: are\: \: given\: \: that\: \: \bar{z}=\cos x-i \sin 2 x$ $$

\sin x-\operatorname{icos} 2 x=\cos x-i \sin 2x Comparing the real and imaginary part, we get

\sin \mathrm{x}=\cos \mathrm{x}$ \: \: and\: \: $\cos 2 \mathrm{x}=\sin 2 \mathrm{x}

tanx=1 and tan2x=1
tan2x=2tanx1tan2x=1
The above value is not satisfied by tan x = 1. Hence no value of x is possible.

Question:36

The real value of α for which the expression (1isinα)1+2isinα is purely real is:
A.(n+1)π2
B.(2n+1)π2
C.nπ
D. None of these

Answer:

The answer is the option (c).
Let z=(1isinα)1+2isinα
=(1isinα)(12isinα)(1+2isinα)(12isinα)=12isinαisinα+2i2sin2α(1)2(2isinα)2=13isinα2sin2α14i2sin2α=(12sin2α)3isinα1+4sin2α3isinα1+4sin2α=0sinα=0α=nπ Hence, c is correct. 

Question:37

If z = x + iy lies in the third quadrant, the z¯zalso lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So,x<0andy<0,z=x+iy
z¯z=xiyx+iy=xiyx+iyxiyxiy=x2+i2y22xyix2i2y2=x2y2x2+y22xyx2+y2i
When z lies in third quadrant then z¯z will also be in third quadrant.
x2y2x2+y2<0 and 2xyx2+y2>0x2y2<0 and 2xy>0x2<y2 and xy>0 So, x<y<0
Hence, b is correct..

Question:38

The value of (z+3)(z¯+3) is equivalent to
A.|z+3|2
B. |z – 3|
C.z2+3
D. None of these

Answer:

The answer is the option (a).
Letz=x+iySo,(z+3)(z¯+3)=(x+iy+3)(xiy+3)
=[(x+3)+iy][(x+3)iy]
=(x+3)2y2i2=(x+3)2+y2
=|x+3+iy|2=|z+3|2

Question:39

If (1+i1i)x=1 , then
A. x=2n+1
B. x=4n
C. x=2n
D.x=4n+1

Answer:

The answer is the option(b).
(1+i1i)x=1((1+i)(1+i)(1i)(1+i))x=1(11+2i1+1)x=1(2i2)x=1(i)x=1
x=4n,nϵN
Hence, the correct option is

A real value of x satisfies the equation 34ix3+4ix=αiβ (α,βR) , if α2+β2=
A. 1
B. –1
C. 2
D. –2

Answer:

The answer is the option (a).

Given that34ix3+4ix=αiβ=34ix3+4ix34ix34ix=αiβ912ix12ix+16i2x2916i2x2=αiβ=916x29+16x224x9+16x2=αiβ.(i)916x29+16x2+24x9+16x2=α+iβ(ii)
Multiplying eqn (i )and (ii )we get
(916x29+16x2)2+(24x9+16x2)2=α2+β2
81+256x4288x2+576x2(9+16x2)2=α2+β2
81+256x4+288x2(9+16x2)2=α2+β2(9+16x2)2(9+16x2)2=α2+β2=1

Question:41

Which of the following is correct for any two complex numbers z1andz2 ?
A. |z1z2|=|z1||z2|
B. arg(z1z2)=arg(z1).Arg(z2)
C. |z1+z2|=|z1|+|z2|
D. |z1+z2||z1||z2|

Answer:

The answer is the option (a).
z1=r1(cosθ1+isinθ2)|z1|=r1z2=r2(cosθ2+isinθ2)|z2|=r2z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]=r1(cosθ1+isinθ1)r2(cosθ2+isinθ2)
=r1r2[(cosθ1cosθ2sinθ1sinθ2)+i(sinθ1cosθ2+cosθ1sinθ2)]=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]|z1z2|=|z1z2|

Hence, option a is correct.

Question:42

The point represented by the complex number 2i is rotated about origin through an angle π/2 in the clockwise direction, the new position of point is:
A. 1+2i
B. 12i
C. 2+i
D. 1+2i

Answer:

The answer is the option (b).
If z rotated through an angle of π/2about the origin in clockwise direction .
Then the new position =z.e(π2)
=(2i).e(π2)
=(2i).[cos(π2)+isin(π2)]
=(2i).(0i)
=12i
Hence, the correct option is (b)

Question:43

Let x,yR, then x+iy is a non-real complex number if:
A. x=0
B.y=0
C. x0
D. y0

Answer:

The answer is the option (d).
x+iy is a non real complex number if y0,

Hence, d is correct

Question:44

If a+ib=c+id, then
A. a2+c2=0
B. b2+c2=0
C. b2+d2=0
D. a2+b2=c2+d2

Answer:

The answer is the option (d).
a+ib=c+id
|a+ib|=|c+id|
a2+b2=c2+d2
Squaring both sides, a2+b2=c2+d2
Hence, d is correct

Question:45

The complex number z which satisfies the condition |i+ziz|=1 lies on
A. circle x2+y2=1
B. the x-axis
C. the y-axis
D. the line x+y=1

Answer:

The answer is the option (b).
 Let z=x+iy|i+x+iyixyi|=1|x(y+1)ix(y1)i|=1|x+(y+1)i|=|x(y1)i|x2+(y+1)2=x2+(y1)2x2+(y+1)2=x2+(y1)2y2+2y+1=y22y+12y=2yy=0;xaxis

Hence, b is correct

Question:46

If z is a complex number, then
A.|z2|>|z|2
B. |z2|equals|z|2
C.|z2|<|z|2
D. |z2||z|2

Answer:

. The answer is the option (b).
Letz=x+iy,|z|2=x2+y2
Now,z2=x2+y2i2+2xyi
z2=x2y2+2xyi
|z|2=(x2y2)2+(2xy)2
=x4+y4+2x2y2=(x2+y2)2
|z|2=x2+y2=|z|2

Hence, b is correct

Question:47

|z1+z2|=|z1|+|z2| is possible if
A. z1=z1¯

B. z2=z1
C. arg(z1)=arg(z2)
D. |z1|=|z2|

Answer:

The answer is the option (c).
Since, |z1+z2|=|z1|+|z2|
|z1+z2|=|r1(cosθ1+isinθ1)+r2(cosθ2+isinθ2)|
=r12cos2θ1+r22cos2θ2+2r1r2cosθ1cosθ2+r12sin2θ1+r22sin2θ2+2r1r2sinθ1sinθ2
So, =r12+r22+2r1r2cos(θ1θ2)|z1+z2|=|z1|+|z2|r12+r22+2r1r2cos(θ1θ2)=r1+r2
Squaring both sides,
r12+r22+2r1r2cos(θ1θ2)=r12+r22+2r1r22r1r22r1r2cos(θ1θ2)=01cos(θ1θ2)=0cos(θ1θ2)=1θ1θ2=0θ1=θ2soarg(z1)=arg(z2)

Question:48

The real value of θ for which the expression 1+icosθ12icosθ is a real number is:
A. nπ+π4


B. nπ+(1)nπ4
C. 2nπ+π2
D. none of these

Answer:

Let \: \: z=\frac{1+i \cos \theta}{1-2 i \cos \theta} =\frac{1+i \cos \theta}{1-2 i \cos \theta} * \frac{1+2 i \cos \theta}{1+2 i \cos \theta} On solving we get
1+3icosθ2cos2θ14i2cos2θ=1+3icosθ2cos2θ1+4cos2θ
=(12cos2θ)+3icosθ1+4cos2θ
If z is real number then
3cosθ1+4cos2θ=0
3cosθ=0cosθ=0
\theta=\frac{(2 n+1) \pi}{2}, n \in N

Question:49

The value of arg(x) when x<0is:
A. 0
B. π2

C. π
D. none of these

Answer:

The answer is the option (c).
Letz=x+0iandx<0|z|=(1)2+(0)2=1,x<0

Since, the point (-x,0)lies on the negative side of the real axis , principal argument (z)=π

Hence, option c is correct

Question:50

If f(z)=7z1z2where z=1+2i, then |f(z)| is
A. |z|2


B. |z|
C. 2|z|
D. none of these

Answer:

The answer is the option (a).
Given that z=1+2i|z|=(1)2+(2)2=5
Now f(z)=7z1z2
=7(1+2i)1(1+2i)2=712i114i24i=62i44i=3i22i
=3i22i2+2i2+2i=6+6i2i2i244i=6+4i+24+4=8+4i8=1+12i
f(z)=(1)2+(12)2=1+14=52=|z|2

Hence, option a is correct

Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5

Various laws need to be followed and derived by a mathematician when it comes to multiplying the two complex numbers. All of them are named over the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axis to form an argand or complex plane.

Students, with the help of NCERT Exemplar Class 11 Maths solutions chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.

All the concepts have been covered in NCERT Exemplar solutions for Class 11 Maths chapter 5. By using NCERT Exemplar Class 11 Maths chapter 5 solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.

Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5

  • Introduction
  • Complex Numbers
  • Algebra of Complex Numbers
  • Addition of two complex numbers
  • Difference of two complex numbers
  • Multiplication of two complex numbers
  • Division of two complex number
  • Power of i
  • The square roots of a negative real number
  • Identities
  • The Modulus and the Conjugate of a Complex Number
  • Argand Plane and Polar Representation
  • polar representation of a complex number
  • Quadratic Equations

What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?

Belonging to the algebraic part of the subject, it is always fun for the students to solve these problems, once they are clear with the concept. NCERT Exemplar Class 11 Maths chapter 5 solutions will help in understanding the said concept and give examples in how to apply them to questions of any kind.

Other than that, in geometric applications, complex numbers are also very useful, for students in their near future. Students, through NCERT Exemplar Class 11 Maths solutions chapter 5, will also learn graphs in an advanced manner with newer placements and some difficult problems, which would make them sharper and clever.

NCERT Solutions for Class 11 Mathematics Chapters

Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

· Class 11 Maths NCERT Exemplar solutions chapter 5 has detailed that complex numbers, their operations, square root, power, identities, representation of complex numbers and quadratic equations are important topics which students should pay extra attention to.

· In NCERT Exemplar Class 11 Maths chapter 5 solutions, students will also learn about quadratic equations. A quadratic equation is an equation where rearrangement of a particular equation is done via a standard form, where x is the unknown number, whereas, a, b and c, are known numbers. An equation with four numbers is said to be quadratic equations.

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Frequently Asked Questions (FAQs)

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Yes, our experts have solved every question in the most detail-oriented way in NCERT Exemplar Class 11 Maths solutions Chapter 5. 

3. Which topics are added in this chapter?

In this chapter, the complex numbers are explained, their properties, operations on complex numbers along with quadratic equations.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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